7. Limits by algebraic simplification
Limits by algebraic simplification
Factor and cancel method
The substitution rule (see 6.1) cannot be used to evaluate limx→a f (x) if a is not in the
domain of the function f (for instance, if it produces a zero in the denominator).
Combining fractions method
Rationalization method
In this section, we get three methods for evaluating limits when substitution fails. Each
method involves an algebraic simplification.
7.1. Factor and cancel method
When substitution produces a zero in both denominator and numerator of a fraction, it
sometimes works to factor and cancel.
7.1.1
Example
x2 + 3x − 4
.
x→−4
x+4
Evaluate lim
Solution First, due to the zero produced in the denominator, the substitution rule cannot
be used. Since the numerator becomes zero as well, we try to factor and cancel:
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x2 + 3x − 4
(x − 1)(x + 4)
= lim
= lim x − 1 = −5.
x→−4
x→−4
x→−4
x+4
x+4
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The reason that the next to the last equality is valid is that the fraction equals x − 1 except
when x = −4, and so the limits of these two functions are the same (since the limit is not
concerned with the behavior of the function right at −4).
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7.2. Combining fractions method
Limits by algebraic simplification
Factor and cancel method
When the expression you are trying to find the limit of contains a sum or difference of fractions, it sometimes works to combine the fractions. This requires a common denominator,
which can always be obtained by just multiplying the two denominators together, though
it usually simplifies computations if a least common denominator is used.
7.2.1
Example
Evaluate lim
x→2
1
1
−
4x − 8 x2 − 4
Combining fractions method
Rationalization method
.
Solution First, due to the zero produced in either one of the denominators, the substitution
rule cannot be used. We combine the fractions by getting a common denominator and then
cancel factors to get a function for which the substitution rule is applicable:
1
1
1
1
− 2
= lim
−
lim
x→2 4(x − 2)
x→2 4x − 8
x −4
(x + 2)(x − 2)
x+2
4
= lim
−
x→2 4(x + 2)(x − 2)
4(x + 2)(x − 2)
(x + 2) − 4
x−2
= lim
= lim
x→2 4(x + 2)(x − 2)
x→2 4(x + 2)(x − 2)
1
1
=
.
= lim
x→2 4(x + 2)
16
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7.3. Rationalization method
Limits by algebraic simplification
Factor and cancel method
When the expression you are trying to find the limit of is a fraction involving a square root,
it sometimes works to rationalize.
The reader probably already knows how to rationalize a numerical denominator:
√
2
2
5−3
√
=√
·√
5+3
5+3
5−3
√
2 5−6
= √ 2
5 − 32
√
√
3− 5
2 5−6
=
.
=
5−9
2
The first step is to multiply numerator and denominator by the expression you are trying
to rationalize except with the opposite sign (the equality sign is valid since this is merely
multiplication by 1). The remaining steps involve simplification.
What makes this process work is the identity (a + b)(a − b) = a2 − b2 . When the left-hand
side of this equation is expanded, the middle terms cancel, yielding the right-hand side.
This identity (written in the reverse order) is the factorization of a difference of two squares.
In the case of limits, the fraction will involve a variable, but the process is the same. Also,
rationalizing the numerator is sometimes called for. Finally, one should use the rationalization method only if substitution produces a zero in both denominator and numerator.
√
7.3.1
Example
Solution
Evaluate lim
x→0
Combining fractions method
Rationalization method
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1+x−1
.
x
First, due to the zero produced in the denominator, the substitution rule cannot
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be used. Since the numerator becomes zero as well, we use the rationalization method to
get a function for which the substitution rule is applicable:
√
√
√
1+x−1
1+x−1
1+x+1
lim
= lim
·√
x→0
x→0
x
x
1+x+1
(1 + x) − 1
1
1
= lim √
= lim √
=
x→0 x( 1 + x + 1)
x→0
2
1+x+1
Limits by algebraic simplification
Factor and cancel method
Combining fractions method
Rationalization method
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7 – Exercises
Limits by algebraic simplification
Factor and cancel method
Combining fractions method
7–1
x2 − 4x
Evaluate lim 2
.
x→4 x − 3x − 4
7–2
Evaluate lim
x2 + x − 12
.
x→3
x+6
√
7–3
7–4
Rationalization method
Evaluate lim
x→4+
Evaluate
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x−2
.
4−x
lim√ sec
x→− 2
−1
x.
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7–5
Evaluate lim
x→0
1
5
−
.
x x2 + 5x
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